A piston-cylinder device contains 25g of steam that is maintained at a constant pressure of 300kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2A for 5min from a 120-V source. At the same time, heat loss of 3.7kJ occurs. Assume steam to behave like ideal gas. Choose the correct statement(s):
$[T a k e : R = \frac{25}{3} J / m o l - K, C_{v} = 4 R, C_{P} = 5 R f o r s t e a m]$

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The rise in temperature of steam is $60.48 ° C$

The increase in volume of steam is $\frac{700}{3} \times 10^{- 5} m^{3}$

Work done by steam on the piston is 700 J

Internal energy of steam falls by 2800 J during the process.

answer is A, B, C.

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Detailed Solution

$N o . o f m o l e s = \frac{25}{18} m o l e s$

$P_{1} = 300 \times 10^{3} N / m^{2} Q_{s o u r c e} = 0.2 \times 120 \times 720 = 7200 J Δ Q_{u s e f u l} = 7200 - 3700 = 3500 Δ Q = Δ U + Δ W P r o c e s s i s i s o b a r i c Δ Q = n C_{p} Δ T 3500 = \frac{25}{18} \times (1 + 4) \frac{25}{3} \times Δ T Δ T = 60.48 ° C Δ U = 2800 J (d U = n C v d T) Δ W = 700 = P \times Δ V Δ V = \frac{700}{3} \times 10^{- 5} m^{3}$